Let’s solve the discrete log problem $2^x\equiv 71(\mathrm{m}\mathrm{o}\mathrm{d}131)$ by the Baby Step-Giant Step method of Subsection 10.2.2. We take $N=12$ since $N^2>p-1=130$ and we form two lists. The first is $2^j(\text{mod}1)31$ for $0\le j\le 11$:

>> for k=0:11;z=[k, powermod(2,k,131)];disp(z);end;

0 1
1 2
2 4
3 8
4 16
5 32
6 64
7 128
8 125
9 119
10 107
11 83

The second is $71\cdot 2^{-j}(\text{mod}1)31$ for $0\le j\le 11$:

>> for k=0:11;z=[k, mod(71*invmodn(powermod(2,12*k,131),131),131)]; disp(z);end;

0 71
1 17
2 124 
3 26 
4 128 
5 86 
6 111 
7 93 
8 85 
9 96 
10 130 
11 116

The number 128 is on both lists, so we see that $2^7\equiv 71\cdot 2^{-12\ast 4}(\text{mod}131)$. Therefore,